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The sequence $4 ; 9 ; x ; 37 ; ...$ is a quadratic sequence.

1. Calculate $x$
2. Hence, or otherwise, determine the nth term of the sequence
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1. 1st difference= $5,x-9,37-x,...$

2nd difference= $x-9-5, 37-x-(x-9),...$

since 2nd difference is constant,

$x-9-5 = 37-x-(x-9)$

$x-14=37-2x+9$

$3x= 60$

$x=20$

2.  $2a= x-9-5 = 20-9-5= 6$

$2a= 6$

$a= 3$

$3a+b=5$

$b= 5-3(3)$

$b= -4$

$a+b+c= 4$

$c= 4-3-(-4)$

$c= 5$

Therefore nth term = $3n^2 -4n +5$
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